Prob. 2, Sec. 2.8 in Kreyszig's functional analysis text: What is the norm of these bounded linear functionals on $C[a,b]$? Let $C[a,b]$ denote the normed space of all the continuous (real or complex-valued) functions defined (and continuous!) on the closed interval $[a,b]$ on the real line, where $a, b \in \mathbb{R}$ and $a < b$, with the norm given by
$$\Vert x \Vert \colon= \max_{t \in [a,b]} \left\vert x(t) \right\vert \ \mbox{ for all } \ x \in C[a,b].$$ 
Let $y_0$ be a given element of $C[a,b]$, and let $\alpha$ and $\beta$ be some given scalars. 
Let the functionals $f_1$ and $f_2$ be defined as follows: 
$$f_1(x) \colon= \int_a^b \ x(t) y_0(t) \ \mathrm{d} t \ \mbox{ for all } \ x \in C[a,b],$$
$$f_2(x) \colon= \alpha x(a) + \beta x(b) \ \mbox{ for all } \ x \in C[a,b].$$ Then both $f_1$ and $f_2$ are bounded linear functionals on $C[a,b]$. Moreover, we have for all $x \in C[a,b]$
$$
\begin{align}
\left\vert f_1(x) \right\vert &= \left\vert \int_a^b \ x(t) y_0(t) \ \mathrm{d} t \ \right\vert \\
&\leq \left( \int_a^b \left\vert y_0(t) \right\vert \ \mathrm{d} t \right) \Vert x \Vert,
\end{align}
$$
showing that $$\Vert f_1 \Vert \leq \int_a^b \left\vert y_0(t) \right\vert \ \mathrm{d} t.$$
And, 
$$ \Vert f_2 \Vert \leq \vert \alpha \vert + \vert \beta \vert. $$
What next?
 A: Ideally you need to find points where these suprema are attained. This is possible for $f_2$ as $\exists x \in C[a,b]$ such that $\|x\| = 1$
$$
x(a) = \text{sgn}(\alpha) \text{ and } x(b) = \text{sgn}(\beta)
$$
where $\text{sgn}(z) = z/|z|$ (This follows from Urysohn's lemma, if you like, although one can just draw the graph of such a function on $[a,b]$). Now for this $x$, one has
$$
f_2(x) = |\alpha| + |\beta|
$$
so your inequality becomes an equality.
For $f_1$ it gets more complicated as the function $t\mapsto \text{sgn}(y_0(t))$ may not be continuous as $y_0$ may have zeroes. To fix this, for each $\epsilon > 0$, consider $x_{\epsilon} \in C[a,b]$ by
$$
x_{\epsilon}(t) = \frac{\overline{y_0(t)}}{|y_0(t)| + \epsilon}
$$
so that if $M := \int_a^b |y_0(t)|dt$, then
$$
M - \epsilon(b-a) = \int_a^b (|y_0(t)| - \epsilon) dt = \int_a^b \frac{|y_0(t)|^2 - \epsilon^2}{|y_0(t)| + \epsilon} \leq \int_a^b y_0(t) x_{\epsilon}(t)dt
$$
Hence
$$
M - \epsilon(b-a) \leq f_1(x_{\epsilon}) \leq \|f_1\|
$$
since $\|x_{\epsilon}\| \leq 1$. This is true for all $\epsilon > 0$, so one gets $M \leq \|f_1\|$, proving that $\|f_1\| = M$ as you have guessed.
